To solve these problems, we can use the kinematic equations of motion. The first equation is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Let's calculate the velocity 15 seconds after the train's departure, given that it has an acceleration of 0.3 m/s^2.
First, we need to determine the initial velocity (u) of the train, which is 0 m/s since it starts from rest. The acceleration (a) is given as 0.3 m/s^2, and the time (t) is 15 seconds.
Plugging these values into the formula:
v = u + at
v = 0 + 0.3 * 15
v = 0 + 4.5
v = 4.5 m/s
So, the velocity of the train 15 seconds after its departure is 4.5 m/s.
To calculate the time taken to reach 30 km/hr (kilometers per hour), we need to convert it to meters per second (m/s).
1 km/h = 1000 m/3600 s = 5/18 m/s
So, 30 km/h is equivalent to (30 * 5/18) m/s.
Now, let's calculate the time taken to reach this velocity.
v = u + at
Given:
u = 0 m/s (since the train starts from rest)
a = 0.3 m/s^2 (the same acceleration as before)
v = 30 * 5/18 m/s (converted velocity)
Plugging in these values and solving for t:
30 * 5/18 = 0 + 0.3t
(30 * 5)/(18 * 0.3) = t
t ≈ 27.78 s
Therefore, it takes approximately 27.78 seconds for the train to reach a velocity of 30 km/h.
To calculate the distance covered by the train, we can use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
Given:
u = 0 m/s (initial velocity)
a = 0.3 m/s^2 (acceleration)
t = 27.78 s (time taken to reach the velocity)
Plugging in the values:
s = (0 * 27.78) + (1/2) * 0.3 * (27.78)^2
s = 0 + 0.5 * 0.3 * 27.78^2
s ≈ 104.15 m
Therefore, the train covers approximately 104.15 meters during the time it takes to reach a velocity of 30 km/h.